### A body moving with constant acceleration | travels the distance 3 metre and 8 metre respectively in 1 second and 2 second | calculate the initial velocity and the acceleration of body | Sum Motion in one Dimension - Physics class 9

Problem : A body moving with constant acceleration travels the distance 3 metre and 8 metre respectively in one second 2 second. Calculate the 1) initial velocity and 2) the acceleration of body.
Data given:

Distance(s1) = 3m

Time taken (t1) to travel 3 m =1 second

Distance(s2) = 8m

Time taken (t2) to travel 8 m =2 second

Initial velocity (u)= ?

Acceleration (a) = ?

Case 1:We know

s = ut + 1/2 a. t2

Or s1 = ut1 + 1/2 a. t12

or 3 = u . 1 + 1/2 .a . 12

or 3 = u + a/2 -------Eqn. 1 Case 2: s2 = ut2 + 1/2 a. t22

or 8 = u . 2 + 1/2 .a . 22

or 8 = 2u + 2a

or 8/2 = (2u + 2a)/2

or 4 = u + a -------Eqn. 2

Subracting Eqn. 1 from Eqn. 2 (Eqn. 2 - Eqn. 1) We have:

4 - 3 = (u + a) - (u + a/2)

Or 1 = u + a - u - a/2

Or 1 = a - a/2

Or 1 = (2a - a)/2

Or 1 = a /2

Or a = 2 m/s2

Putting the value of a in Eqn. 1 we have:

3 = u + 2/2

3 = u + 1

Or u = 3 - 1

Or u = 2 m/s

∴ Initial velocity (u)= 2 m/s

Acceleration (a) = 2 m/s2Answer

Distance(s1) = 3m

Time taken (t1) to travel 3 m =1 second

Distance(s2) = 8m

Time taken (t2) to travel 8 m =2 second

Initial velocity (u)= ?

Acceleration (a) = ?

Case 1:We know

s = ut + 1/2 a. t2

Or s1 = ut1 + 1/2 a. t12

or 3 = u . 1 + 1/2 .a . 12

or 3 = u + a/2 -------Eqn. 1 Case 2: s2 = ut2 + 1/2 a. t22

or 8 = u . 2 + 1/2 .a . 22

or 8 = 2u + 2a

or 8/2 = (2u + 2a)/2

or 4 = u + a -------Eqn. 2

Subracting Eqn. 1 from Eqn. 2 (Eqn. 2 - Eqn. 1) We have:

4 - 3 = (u + a) - (u + a/2)

Or 1 = u + a - u - a/2

Or 1 = a - a/2

Or 1 = (2a - a)/2

Or 1 = a /2

Or a = 2 m/s2

Putting the value of a in Eqn. 1 we have:

3 = u + 2/2

3 = u + 1

Or u = 3 - 1

Or u = 2 m/s

∴ Initial velocity (u)= 2 m/s

Acceleration (a) = 2 m/s2Answer